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Derivative of Field-Operator and Vector-Potential

  1. Jan 15, 2014 #1
    Hello,

    I regard field-operators, whereby [itex]\Psi[/itex] is a fermionic-annihilation-operator and

    [itex]\vec{A(\vec{r})}[/itex] is an electromagnetic-vector-potential.

    Is it possible to do the following step?


    $$
    \nabla \vec{A} \Psi =\Psi \nabla \vec{A} + \vec{A}\nabla\Psi
    $$

    And if its correct, why?

    Thx
    Abby
     
  2. jcsd
  3. Jan 15, 2014 #2
    Ok I think I got a Solution.
    If I consider Coulomb Gauge

    $$
    \nabla \vec{A}=0
    $$
    and I can write
    $$
    \nabla \vec{A} \Psi = \vec{A}\nabla\Psi
    $$
     
  4. Jan 17, 2014 #3

    Avodyne

    User Avatar
    Science Advisor

    $$
    \nabla\cdot( \vec{A} \Psi) =(\nabla\cdot \vec{A})\Psi + \vec{A}\cdot\nabla\Psi
    $$
    by the product rule for derivatives. If you like, you can move [itex]\Psi[/itex] to the left in the first term, since [itex]\Psi[/itex] and [itex]\vec A[/itex] commute.
     
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